Let $X_i$ be independent random variable such that each $X_i$ is symmetric about 0 and Var($X_i) = 2i -1$, for $i \geq 1$. Then $$\lim_{n \mapsto \infty} P(X_1 + \ldots + X_n > n {\rm log}\;n) = ?$$ Options:
a) doesn't exists
b) equals 1/2
c) equals 1
d) equals 0
Right answer b)
My approach:
Since question involved inequality, I thought I could apply Chebychev inequality. But before that I made few observations,
1) $E(X_i) = 0$ as it is symmetric about $0$.
2) $P(\mid X \mid \geq \epsilon) = 2 P(X \geq \epsilon)$ because of symmetry around $0$.
Let $S = X_1 + \ldots + X_n$. So I wrote the equation $$2 \times P(S \geq k\sigma )\leq 1/ k^2$$ Identifying $n$ log $n = k \sigma$ where $\sigma = $Var$(S)$, my answer should be $\sigma^2 / (2 \times n$ log $n)^2$. To calculate $\sigma$ observe that $X_i$ are independent and hence $\sigma = \sum_{i=1}^{n}(2i -1)$ which is $n^2$. Hence the final expression should be $$\lim_{n \mapsto \infty}\big(n^2/(2 \times n {\rm log} n) \big)^2 = 0$$ which is different from the answer given. Examining the expression again, there is $1/2$ term present which would mean that maybe I made mistake in caluclating the limits. So can anyone help me to sort this out?
Reference: CSIR NET DEC 2015 Paper Code A Q.no. 53